Statistical Genomics Zhiwu Zhang Washington State University Lecture 4: Statistical inference

 Homework1, due Feb 3, Wednesday, 3:10PM Administration

 X2 test on contingency table  Empirical null distribution  X2 test on variance  t test  Hypothesis test  two types of error  Power Outline

TransgeneticNon transgeneticSUM Herbicide35540 No herbicide SUM Observed and expected frequency TransgeneticNon transgeneticSUM Herbicide No herbicide SUM

 Poisson distribution: Mean=Var=Expected  (Observed-Expected)/Sqrt(Expected) ~ N(0,1)  SUM(Observed-Expected) 2 / Expected ~ X 2 (df)  df=number of independent cells Approximate Distributions

TransgeneticNon transgeneticSUM Herbicide35540 No herbicide SUM Observed and expected frequency TransgeneticNon transgeneticSUM Herbicide No herbicide SUM /28+49/12+49/42+49/18=9.72

Distribution of x2(1) Observed 9.72 P<1% 99% percentile 6.97 par(mfrow=c(2,2),mar = c(3,4,1,1)) x=rchisq(k,1) d=density(x) plot(x) plot(d) hist(x) plot(ecdf(x)) quantile(x,.99)

 A sample has mean of and variance of  The sample has 10 observations  Q1: What is the probability that the sample was from a normal distribution with variance of 25?  Q2: What is the probability that the sample was from a normal distribution with mean of 100? Tests on samples

 Empirical solution:  Sample ten observations from a normal distribution with variance of 25.  Calculate observed variance.  Repeat the sampling and get null distribution of the sample variances  Find percentile of observed variance on the null distribution Q1: distribution with variance of 25

x=replicate(10000, {s=rnorm(10,0,5) var=var(s) }) Observed P>25% 75% percentile 31.6 > length(x[x>27.82])/10000 [1] par(mfrow=c(2,2),mar = c(3,4,1,1)) d=density(x) plot(x) plot(d) hist(x) plot(ecdf(x)) quantile(x,.75)

 Theoretical solution: Q1: distribution with variance of 25 v=(10-1)*27.82/25= > 1-pchisq(10.026,9) [1] vs from empirical

Q2: distribution with mean of 100  Empirical solution  Sample ten observations from N(100, 25)  Calculate mean  Repeat the process 10,000 times  Null distribution of of the 10,000 means  Determine the percentile of testing mean (103.6) on the null distribution

Q2: distribution with mean of 100 x=replicate(10000, {s=rnorm(10,100,5) m=mean(s) }) Observed %<P<5% 95% percentile > length(x[x>103.6])/10000 [1] par(mfrow=c(2,2),mar = c(3,4,1,1)) d=density(x) plot(x) plot(d) hist(x) plot(ecdf(x)) quantile(x,.95) quantile(x,.99) 99% percentile 102.6

t test

T=( )/(5/sqrt(10)) P=1-pt(T,9) c(T,P) Under 5% of threshold, reject the hypothesis that the sample was from a distribution with mean of 100

F test

 Null hypothesis (H0): Initial assumption  Alternative hypothesis (Ha): Opposite to the assumption  Find the probability of H0  If the probability is too low (e.g. 5%), reject Ho and accept Ha  Otherwise, accept Ho Hypothesis test

 Type I error: Reject true H0, False positive, the probability is the threshold used, e.g. α=5%  Type II error: Accept false H0, false negative, β  Power: Probability to reject false H0, (1-β) Two types of errors and power

TestH0 is TrueHo is False Positive (reject H0) False positive Type I: α Power=1-β Negative (Accept H0) Specificity=1-α False negative Type II: β Sum100% Summary

Highlight  X2 test on contingency table  Empirical null distribution  X2 test on variance  t test  Hypothesis test  two types of error  Power